Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const long long INF = 1e18;
const long double EPS = 1e-12;
int a[105][105];
int b[105][105];
vector<pair<pair<int, int>, int> > vec[505];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < (int)n; i++) {
    for (int j = 0; j < (int)m; j++) {
      cin >> a[i][j];
    }
  }
  int mx = (int)(1e9);
  for (int i = 0; i < (int)m; i++) {
    mx = min(mx, a[0][i]);
  }
  int ans = (int)(2e9);
  int idx = 0;
  for (int di = 0; di < (int)mx + 1; di++) {
    int net = 0;
    int flag = 0;
    net += di;
    if (di != 0) vec[di].push_back(make_pair(make_pair(di, 0), 0));
    for (int i = 0; i < (int)n; i++) {
      for (int j = 0; j < (int)m; j++) {
        b[i][j] = 0;
      }
    }
    for (int j = 0; j < (int)m; j++) {
      b[0][j] += di;
    }
    for (int i = 0; i < (int)m; i++) {
      int dif = a[0][i] - di;
      net += dif;
      if (dif != 0) vec[di].push_back(make_pair(make_pair(dif, i), 1));
      for (int j = (int)1; j < (int)n; j++) {
        b[j][i] += dif;
      }
    }
    for (int i = (int)1; i < (int)n; i++) {
      int dif = a[i][0] - b[i][0];
      if (dif < 0) flag = 1;
      net += dif;
      if (dif != 0) vec[di].push_back(make_pair(make_pair(dif, i), 0));
      for (int j = 0; j < (int)m; j++) {
        b[i][j] += dif;
        if (b[i][j] != a[i][j]) flag = 1;
      }
    }
    if (flag == 0) {
      if (ans > net) {
        ans = net;
        idx = di;
      }
    }
  }
  if (ans != (int)(2e9)) {
    cout << ans << endl;
    for (int i = 0; i < (int)vec[idx].size(); i++) {
      int val = vec[idx][i].first.first;
      int id = vec[idx][i].first.second + 1;
      if (vec[idx][i].second == 0) {
        for (int j = 0; j < (int)val; j++) {
          cout << "row " << id << endl;
        }
      } else {
        for (int j = 0; j < (int)val; j++) {
          cout << "col " << id << endl;
        }
      }
    }
  } else {
    cout << "-1" << endl;
    ;
  }
  return 0;
}